# Tag Info

1

The cutoff frequency of an RC filter is determined by the total resistance and capacitance at the output. With R = 100 Ω and C = 3.62 μF the cutoff frequency is ~440 Hz. When you put 100 Ω across the capacitor the total (Thevenin equivalent) resistance is now 50 Ω, so the cutoff frequency rises to ~880 Hz. You see less dynamic error because the filter is ...

1

T=RC Case 1 RC = 100 * 3.62 uF f-3dB ~ 440 Hz A(50Hz) = -0.05 dB Case 2 RC = 50 * 3.62 uF . f-3dB ~ 880 Hz thus less Attenuation at 50 Hz. This is not really an anti-aliasing filter and no specs. no sampling rate was given nor BW ripple, band-reject attenuation and SNR from Nyquist noise. You should give a spec for any design: e.g. attenuation error ...

2

simulate this circuit – Schematic created using CircuitLab Figure 1. (a) Your circuit. (b) The equivalent circuit which makes it clear that you've created a 2:1 potential divider. (c) The Thevenin equivalent has half the signal level feeding in and 50 Ω source resistance. By adding the load you have overloaded the filter to the point of altering its ...

2

If +5 V and -5 V are regulated outputs of 1 or 2 DC/DC converters than I would use a CMF on the power supply input line. If not, then you need two CMF. The first between +5 V and GND. The second between -5 V and GND. If your design is committed to high production volume, then I suggest you to find a Taiwanese manufacturer and buy directly from them. Cost of ...

0

Why are there so few options for common mode chokes with three lines? What is a good topology for this situation when using common mode chokes with two lines? Common mode chokes on two lines by their very nature cancel differential current MMFs from the load and therefore, can be made from much smaller ferrite components. How would these cancel with three ...

0

You are absolutely right in your thinking, for a symmetric filter, the delay is half of the coefficients. You can verify this by thinking how an unit impulse would go through the structure.

0

The time taken for all the samples to go through the filter is n clocks. However this is not what 'latency' means in the context of signal processing. In many other fields, for airplane turnround, or HDD reads for instance, latency is measured to the completion of an operation. In signal processing, we measure the group delay of the signal, usually by ...

3

The LM339 has an open-collector output stage, which makes it's high state output value dependent on many things - none of which are the LM339 itself. When the output is supposed to be high, the voltage level is dependent on VDD5B, R6, R7, R8, R9, the input voltage OUTB and its source impedance, and whatever OUT2 is connected to, its bias point voltage, and ...

0

You suspect the power supply, but it is guilty? Set the micro to log what it's doing via UART and record it with a PC. Let it run for a while and check the logs. You may find something like the micro rebooting just after it did something. That will narrow it down. Next, stick a scope probe on the micro's VCC, and one on the input voltage. Set the trigger to ...

0

Can LC filter at the input of the switching regulator cause problem? Yes they can (more below) but I'm unsure whether it is the root cause of your real problem associated with your communication failure so, I'd keep looking for that. The problem with an LC at the front input side of any regulating circuit is that if the load on the output of the regulator ...

0

I've never had problems with LC filters in front of DC/DC converters even with power hungry loads. Time ago I designed 3 boards. Each board had DC/DC converter (*) and a GSM module which drew pulsed currents of 2 A and pulse width of 577 micro seconds. I had an LC filter just like you in front of the DC/DC converter. L was 151 micro Henry and C was 220 micro ...

2

Your 2nd figure shows that your output levels are of 4.5V. But according to datasheet lm339A is not rail to rail. So your max output should be of 3.4V. By changing the resistor R6 from 3K to 10K will solve the problem.

-2

The transfer function of a 5th order Low-Pass Filter with a load resistance RL and a gain Resistance Rm: is: H = (RL*Rm)/(K.s^5+L.s^4+M.s^3+N.s^2+P.s+Q) with: K = C0*C1*C2*C3*Cm*R0*R1*R2*R3*Rg*RL*Rm; L = C0*C1*C2*C3*R0*R1*R2*R3*Rg*RL+C0*C1*C2*C3*R0*R1*R2*R3*RL*Rm+C0*C1*C2*C3*R1*R2*R3*Rg*RL*Rm+C0*C1*C2*Cm*R0*R1*R2*R3*Rg*Rm+C0*C1*C2*Cm*R0*R1*R2*Rg*RL*Rm+C0*C1*...

1

I believe the combination of your Q-boosted Delyiannis-Friend bandpass circuit saturating the output stage of the built-in OP284 model is causing the issues. I built your circuit and was able to get it to run the full 1ms by either using UniversalOpamp2 instead of OP284, using the alternate solver instead of normal solver, or setting .options abstol=1u ...

2

Check the reltol/voltol etc. options. Either with the .option command, or behind the hammer icon (4th icon in the bar). Make the tolerances higher for faster simulation speeds while sacrificing accuracy. You have a floating node behind a reactive element at C5. It's good practice to avoid floating nodes in a circuit. I guess this node is where the output ...

-1

Ingenious simplicity. I am sorry but I can not agree with such offensive classifications regarding this circuit - "crappy", "nasty", etc. If it really was, it would not exist perhaps more than a century. Instead, I would say this is another example of what Don Lancaster has called "elegant simplicity." Moreover, I would say &...

2

I can only answer the part of the reflow-soldering, I lack some knowledge about impedance. Joining the pads that way will definitely cause problems during reflow soldering. The problem will be tomb-stoning. This is when a component is standing upright (feet up in the air) on the PCB during reflow soldering. The components will float to the centre of the pad ...

1

Well one of the answers above was mostly right. It is the most basic of a single phase ac to dc full wave rectifier. The cap is in there to smooth the ripple and the res is to discharge the cap when power is disconnected. And the above answer is spot on about never assuming someone would never put their little piggies in where they don't blong and 470uF is ...

0

"what kind of simple analog circuit would turn 1 MHz into 1 kHz and 1.5 MHz into 1.5 kHz?" Analog? None. To compress frequency scale, only "analog" solution is to expand time. The unit of frequency is Hz, which is 1/s unit. Another problem is the amount of this "smog". We are talking about packing tons of information to 1:1000 ...

5

I like the other ideas suggested, here, but just because it isn't stressed much in the answers, bleed resistors seem quite likely. The one in parallel with the LEDs makes sense to me aside from the turnoff and other reasons mentioned could be because resistors are generally more reliable than LEDs and also the most likely time someone will cut the power to ...

24

That is a de facto circuit for cheap generic mains powered LED lamps. The capacitor is an essential part, the circuit is called a "capacitive dropper" which is used to bring down the voltage at modest current level to power the LEDs. The 220k resistor is necessary to keep the LEDs turned off, so that they do not glow faintly due to capacitive ...

6

That's a cheap and nasty "capacitive-dropper" low-voltage PSU. Note that the 470 kΩ resistor is too high a value to drive the LEDs. Instead the current is provided through the 105J capacitor which, presumably, has the right impedance at 50 Hz (220 V land) to power the LEDs. The resistor will discharge the capacitor when unplugged. Since voltage ...

11

Not sure why they put the 220K there. It's not essential to the operation- it will make the LEDs go completely dark a bit faster when power is removed. As others have mentioned, it will help prevent a faint glow from the LEDs caused by capacitive coupling to the wires (say there is a long run of wires to the on/off switch, a tiny bit of current will flow ...

2

Well, the transfer function of this circuit is given by: $$\mathscr{H}\left(\text{s}\right):=\frac{\text{v}_\text{out}\left(\text{s}\right)}{\text{v}_\text{in}\left(\text{s}\right)}=1-\frac{1}{2+\text{CRs}}\tag1$$ When working with sinusoidal signals, we can use $\text{s}:=\text{j}\omega$: \underline{\mathscr{H}}\left(\text{j}\omega\right)=1-\frac{1}{2+\...

0

Solution 1: Increase the PWM frequency on whatever generates it, see if it fixes the problem. If your electromagnet is a coil, it should be able to smooth current... Solution 2: Replace your on/off constant current LED driver with one that supports analog dimming, for example this one or this one. Give it a control voltage, you get the current you want. If ...

0

"Would a parallel array of capacitors lower the ESR and heating?" Yes, it will. It's a common thing putting in parallel, for example, 2 electrolytic capacitors to lower the overall ESR and the temperature.

2

It looks like a high pass shelving filter as Andy says. Whether it is or not depends on the load impedance. If that has a capacitive component it may be a compensated attenuator, such as found in a 10x scope probe.

2

Which type of filter is it? In the field of audio processing and tone controls it's a type of shelving filter: - Ignore the red line - that is also a shelving filter but working at bass frequencies. Yours is equivalent to a treble shelving filter (blue line). Picture from here. Of course, it needn't be associated with audio; it can be used in many signal ...

1

It's possible you're overthinking this. $s$ is the Laplace notation for the complex frequency, $j2\pi f$, or $j\omega$, and it spans the entire frequency range. When you're saying that $|s|\rightarrow0$ or $|s|\rightarrow\infty$, what you're probably referring to is the frequency, itself, as it approaches DC or as it goes to infinity, not $s$. ...

3

You're headed for a regulatory compliance issue The problem with jacking up Cy is that doing so increases the leakage current from the mains to your appliance's earth grounding lead. Given that the least restrictive condition IEC(UL) 60950 permits for garden-variety devices is 3.5mA of earth leakage, this limits Cy to just under 20nF for a reasonable worst ...

6

It is not correct that for "all filters" the corner or cut-off frequency is defined by the "-3dB point" (magnitude 3 dB down with respect to the maximum). This is only the case for all first-order low- and highpass responses as well as 2nd-order bandpass filters, and for higher-order filters with Butterworth characteristics. For all ...

7

It's not 3dB absolute, it's 3dB down from the peak, or some sort of nominal attenuation. So in your case, where the passband is -3dB, 3dB down is at -6dB. Note that some filters (e.g. Chebychev) have significant passband ripple; if this exceeds 3dB then the "3dB down" figure loses meaning. In that case, or just if it's what matters to the system ...

1

This Sigma Delta converter uses over-sampling to obtain the resolution then decimation to reduce the signal BW which is $f_{ADC} /2$ maximum Nyquist rate. It has a built in SINC filter to notch desired grid e-field noise from unbalanced signals. Analog and digital anti-aliasing A typical choice might be 64x oversampling, in which case the ADC will sample ...

0

what is the relevant sampling frequency to use as a basis for calculating the anti aliasing filter. ... does it depend on the specific data rate I'm using? Yes, it depends upon the specific data rate your are using. See for example this document. The sampling rate to use for determining the cutoff frequency of your anti-aliasing filter is the rate at which ...

2

One of the most important uses of buffers (being them emitter followers, voltage followers or whatever) is that the isolate the stages presenting a stable impedance on both input and output. Since a passive filter is influenced by these impedences (RF people fight with this all the day) simply chaining them has the first one seeing the input of the second ...

0

The TL081 has a GBP (Gain Bandwidth Product) of 4 MHz. Its transfer function, which has 2 poles, do participate to the overall transfer function of your circuit. The TL081 is used in your circuit is a unity gain buffer. That means that the first pole is at 4 MHz and that means that starting from 400 KHz the overall transfer function starts "feeling"...

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